Question: $\begin{aligned} &F(x)=4x^2-5x \\\\ &f(x)=F'(x) \end{aligned}$ $\int_1^4 f(x)\,dx=$
$f$ is the derivative of $F$, which means $F$ is an antiderivative of $f$. Since we know the antiderivative of $f$, we can use the fundamental theorem of calculus: For every function $f$ and its antiderivative $F$, $\int_a^b f(x)\,dx=F(b)-F(a)$. $\begin{aligned} &\phantom{=}\int_1^4 f(x)\,dx \\\\ &=F({4})-F({1}) \\\\ &=[4({4})^2-5({4})]-[4({1})^2-5({1})] \\\\ &=44-(-1) \\\\ &=45 \end{aligned}$ In conclusion, $\int_1^4 f(x)\,dx=45$